Question

Let x be a random variable that represents white blood cell count per cubic milliliter of whole blood. Assume that x has a distribution that is approximately normal, with mean μ = 8100 and estimated standard deviation σ = 2600. A test result of x < 3500 is an indication of leukopenia. This indicates bone marrow depression that may be the result of a viral infection.

(a) What is the probability that, on a single test, x is less than 3500? (Round your answer to four decimal places.) (a) What is the probability that, on a single test, x is less than 3500? (Round your answer to four decimal places.)

(b) Suppose a doctor uses the average x for two tests taken about a week apart. What can we say about the probability distribution of x?

The probability distribution of x is not normal.

The probability distribution of x is approximately normal with μx = 8100 and σx = 1838.48.

The probability distribution of x is approximately normal with μx = 8100 and σx = 1300.00.

The probability distribution of x is approximately normal with μx = 8100 and σx = 2600.

What is the probability of x < 3500? (Round your answer to four decimal places.)

(c) Repeat part (b) for n = 3 tests taken a week apart. (Round your answer to four decimal places.)

(d) Compare your answers to parts (a), (b), and (c).

How did the probabilities change as n increased?

The probabilities decreased as n increased.The probabilities increased as n increased.   

The probabilities stayed the same as n increased.

If a person had x < 3500 based on three tests, what conclusion would you draw as a doctor or a nurse?

It would be a common event for a person to have two or three tests below 3,500 purely by chance.

The person probably does not have leukopenia.It would be an extremely rare event for a person to have two or three tests below 3,500 purely by chance. The person probably has leukopenia.    

It would be an extremely rare event for a person to have two or three tests below 3,500 purely by chance. The person probably does not have leukopenia.

It would be a common event for a person to have two or three tests below 3,500 purely by chance. The person probably has leukopenia.

Answer 1

Solution :

Given that ,

mean = = 8100

standard deviation = = 2600

a) P(x < 3500) = P[(x - ) / < (3500 - 8100) / 2600]

= P(z < -1.77)

Using z table,

= 0.0384

b) n = 2

= = 8100

= / n = 2600/ 2 = 1838.48

The probability distribution of x is approximately normal with μx = 8100 and σx = 1838.48

P( < 3500) = P(( - ) / < (3500 - 8100) / 1838.48)

= P(z < -2.50)

Using z table

= 0.0062

c) n = 3

= = 8100

= / n = 2600/ 3 = 1501.11

The probability distribution of x is approximately normal with μx = 8100 and σx = 1501.11

P( < 3500) = P(( - ) / < (3500 - 8100) / 1501.11)

= P(z < -3.06)

Using z table

= 0.0011

d) The probabilities decreased as n increased.

.It would be an extremely rare event for a person to have two or three tests below 3,500 purely by chance. The person probably has leukopenia.